Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/257

249

DES FONCTIONS.

${\frac {\operatorname {d} ^{3}U}{\operatorname {d} b^{3}}}={\frac {\operatorname {d} ^{2}.z^{2}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a\operatorname {d} b}}={\frac {\operatorname {d} ^{2}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}}}\,;$

et de celle-là

${\frac {\operatorname {d} ^{4}U}{\operatorname {d} b^{4}}}={\frac {\operatorname {d} ^{3}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}\operatorname {d} b}}={\frac {\operatorname {d} ^{3}.z^{4}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{3}}}\,;$

et ainsi de suite ; de manière qu’on aura, en général

${\frac {\operatorname {d} ^{m}U}{\operatorname {d} b^{m}}}={\frac {\operatorname {d} ^{m-1}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{m-1}}}.\qquad$(11)

Une fois parvenu à la formule (11), le développement de $U,$ suivant les puissances de $b$ ne présente plus de difficulté.

Soit encore

$x=\operatorname {f} (a+bz)\,;\qquad x_{1}=\operatorname {f_{1}} (a_{1}+b_{1}z_{1})\,;$

$z,z_{1}$ étant des fonctions quelconques de $x,x_{1}$ sans $a,b,a_{1},b_{1}.$ On trouvera, par la différentiation,

${\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} a}}=\left(1+b{\frac {\operatorname {d} z}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b{\frac {\operatorname {d} z}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f} '(a+bz),\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=\left(b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f_{1}} '(a_{1}+b_{1}z_{1}).\end{aligned}}$

En posant, pour abréger,

$k={\frac {b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}{1-b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}},$

la dernière de ces équations donne